Note that if you find that $(x,y)$ is a saddle point for some specific $x,y$ then since $z$ is periodic in $x$ there are infinitely many saddle points of the form $(x+2\pi n,y).$
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coffeemathNov 19 '12 at 7:12

1 Answer
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You first find the critical points of $z=y \sin x$ by finding the points $(x,y)$ for which both partial derivatives are zero. Denoting the partials as $z_x$ and $z_y$, you have $$z_x=y \cos x,$$ $$z_y=\sin x.$$
If these are both zero then $\sin x = 0$ implying $x=n \pi$ for some integer $n$. But then $\cos x=1,-1$ so from $z_x=0$ you see that $y=0$ at any critical point.

So your critical points are all points of the form $(x,y)=(n \pi, 0).$

The next thing one does for checking for saddle points is to compute the following value at each critical point $(x,y)$:
$$J(x,y)=z_{x,x} \cdot z_{y,y} - z_{x,y} z_{y,x},$$
where the double subscripts denote second partials (partials of partials). If this calculation comes out negative, you know you have a saddle point.

I'll leave the rest to you --- you should find that all the above critical points are saddle points.

Note: $J(x,y)$ above is actually the determinant of the "jacobian" at $(x,y)$, where the jacobian is really a matrix. In general one sees whether the jacobian is positive definite, negative definite, or indefinite, or none of these. That is, if the value $J(x,y)$ happens to be $0$ at a critical point, then nothing can be concluded about whether one has a local max, local min, or saddle point.

@coffeemath: The matrix of second partials of a scalar function is usually called the Hessian and denoted by $H$. The notion of Jacobian (and the letter $J$) is relevant for the first derivative of a map $f:\ {\mathbb R}^n\to{\mathbb R}^m$.
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Christian BlatterNov 21 '12 at 9:32